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Examples
Example:
Consider the IVP
2
x
2
y
00
+ 3
xy
0

3
y
= 5
√
x
y
(1) = 1
,y
0
(1) = 1
.
(
NH
)
Strategy: ﬁnd a FS for the associated ho
mogeneous SOLODE
2
x
2
y
00
+ 3
xy
0

3
y
= 0
,
(
H
)
then, using Variation of Parameters, ﬁnd
a particular solution
Y
(
x
)
of
(
NH
)
, which
will give us the general solution
y
(
x
) =
Y
(
x
) +
c
1
y
1
(
x
) +
c
2
y
2
(
x
)
,
(
G
)
then accommodate the Initial Conditions.
See that
(
H
)
has the solution
y
1
(
x
) =
x
.
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View Full Document Panic! I have forgotten how Reduction
of Order works.
Actually, I do remember
that it uses Abel’s theorem: there is a
second solution
y
2
such that
W
[
y
1
,y
2
](
x
) =
e

R
p
(
x
)
dx
.
Now
y
1
y
0
2

y
2
y
0
1
=
e

R
3
/
2
x dx
=
x

3
/
2
=
x

3
/
2
=
⇒
y
1
y
0
2

y
2
y
0
1
y
2
1
=
x

3
/
2
x
2
=
x

7
/
2
=
⇒
±
y
2
y
1
²
0
=
x

7
/
2
=
⇒
y
2
y
1
=
Z
x

7
/
2
dx
=

2
5
x

5
/
2
2
=
⇒
y
2
(
x
) =

5
2
x

3
/
2
Thus
{
x,x

3
/
2
}
is a FS for
(
H
)
. It has Wronskian
W
[
y
1
,y
2
] =

5
2
x

3
/
2
.
Now ﬁnd a particular solution
Y
of
(
NH
)
.
Panic! I have forgotten how Variation of
Parameters works.
Actually, I remember
that it means looking for a particular so
lution of the form
Y
(
x
) =
v
1
(
x
)
y
1
(
x
) +
v
2
(
x
)
y
2
(
x
)
with variable
parameters
v
1
,v
2
. So I take
some scratch paper and derive it right quick:
3
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View Full Document Y
00
=
v
1
y
00
1
+
v
2
y
00
2
+
v
0
1
y
0
1
+
v
0
2
y
0
2
p
Y
0
=
v
1
p
y
0
1
+
v
2
p
y
0
2
+
p
[
v
0
1
y
1
+
v
0
2
y
2

{z
}
=0
]
q
Y
=
v
1
q
y
1
+
v
2
q
y
2
L
[
Y
] =
v
1
L
[
y
1
] +
v
2
L
[
y
2
]

{z
}
=0
+
v
0
1
y
0
1
+
v
0
2
y
0
2

{z
}
=
g
results in the two linear equations
v
0
1
y
1
+
v
0
2
y
2
= 0
,
v
0
1
y
0
1
+
v
0
2
y
0
2
=
g
.
for
v
0
1
,v
0
2
. They have the solutions
v
0
1
=

y
2
g
W
[
y
1
,y
2
]
and
v
0
2
=
y
1
g
W
[
y
1
,y
2
]
.
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This note was uploaded on 09/29/2009 for the course M 427K taught by Professor Fonken during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Fonken
 Differential Equations, Equations

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